Re: Eric Gisse, your comment was ridiculous
From: The Ghost In The Machine (ewill_at_aurigae.athghost7038suus.net)
Date: 08/20/04
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Date: Fri, 20 Aug 2004 08:00:39 GMT
In sci.math, eleaticus
<eleaticus@bellsouth.net>
wrote
on Wed, 18 Aug 2004 15:41:09 -0500
<CcPUc.503$nW6.227@bignews4.bellsouth.net>:
>
> "The Ghost In The Machine" <ewill@aurigae.athghost7038suus.net> wrote in
> message news:kb7cv1-9u4.ln1@lexi2.athghost7038suus.net...
>
>> Boiled down, the Universe is. :-) We try to explain it using
>> mathematics and/or concepts, and don't always get it right.
>
> Hey Ghost, Nice post.
Thank you.
>
>> However, there is one thing the Universe is constant at: the
>> speed of light. (At least, so goes the theory, and so far
>> I've not seen anything that disproves it.) The Newtonian
>> transformation can be expressed:
>>
>> [1] x' = x-vt
>> t' = t
>
> Thematic of my posts but perhaps trivial in the context of what you have
> below (not sure) is:
> no, no, no.
>
> Not t'=t. Just t. Just as in both galilean and LE just v and not v'=v.
The intent is to represent the viewpoint of two observers.
The first observer uses the coordinate system (x,t).
The second uses (x',t'), when measuring spacetime events.
>
> There is no time transform. This spurious xform is the basis of some
> corrupt attacks on the galilean transforms.
One can contemplate the existence of an identity time transform
and see how far one gets.
>
>> if one assumes observations are done using light waves, and
>> the origin contains a Master Clock or some such.
>
>> Now shoot a lightray from (x,t)=(0,0). The equation of that
>> light ray will be x = ct. If we transform it:
>
>> x' = ct-vt = (c-v)t = (c-v)t'
>
>> Since Michelson-Morley didn't show any variance (we're moving
>> at 10^-4 c, after all), this doesn't work very well.
>
> Michelson-Morley works very well galilean-wise, perfectly in one
> version of c+v: the 'catch-and-throw' or 'intrinsic energy' model
> of light , but not in the 'bounce' or 'kinetic mass/energy' model
> of light. [Do ask.]
>
> To return to what you just said: every ninth grader knows that
> to use their ruler/yardstick when for some reason you aren't
> going to keep the zero mark/end at one end of the length being
> measured, you use the difference of the two marks/coordinates of
> the length's two ends. Say, x0 and x1. Not a problem when setup
> with the origin conveniently placed but otherwise, and when one
> then intends to move the origin or relate to a less propituously
> placed ruler, use the difference expression;
>
> Hence x1 = x0 + ct, (x1-x0)=ct.
>
> Then, x1' = x0' + ct, (x1'-x0') = ct = (x1-x0). Invariance of all equations
> depending on lengths/distances..
>
> Thus, whether the x-origin is well placed or not at time t=0, as long as we
> use x0 as the 'left' end of the measured length/distance, we have no
> problem except with the virulent, vicious onslaught by true believers?
>
> Children know how to do it right. What is the impediment to clear thought by
> Relativists?
Try using that stick to measure a moving car. There are certain
logistical difficulties. :-)
But you do have a point; the stick can either be used by placing
the rod (or other implement) next to the origin and measuring
along the length for an absolute reading, or placing the rod
next to any part of the stick and subtracting the endpoints
for a relative one.
>
> Heck, even psychologists and sociologists have learned to do it right (in a
> manner of speaking).
>
>> If we modify the time definition in what looks like a logical manner:
>>
>> [2] x' = x-vt
>> t' = t-x/c
>
> I dig you being friendly and playful here and below but haven't yet
> extracted both the intellectual motivation and good insight as to how
> respond other than to say: no time transform.
>
> BTW, not directly to do with your post but to Uncle Al's and others'
> diatribes, this riddle:
>
> If we call cats dogs we can then reasonably say lots of erroneous things
> about dogs.
Calling a tail a leg does not make it a leg.
-- attributed to Abraham Lincoln.
The Universe is a capricious place. Without the time transform
one has some difficulties explaining the orbit of Mercury,
for example. There are also issues with atmospheric muons
and the movement of an iron block while irradiated with
a certain frequency of gamma rays, and the subsequent
measurement of the absorption thereof.
I shall have to analyze this transform in more detail; my
algebra seems to be slightly defective. However, it should
be possible to use Minkowski invariance to do what I want;
the basic issue there is that, if two observers see two
events at coordinates (x1,t1) and (x2,t2) (and (x1',t1')
and (x2',t2')), then the distance should be the same, and
the distance in this case is the rather strange-looking
but logical metric (x2-x1)-c^2*(t2-t1).
But simultaneity in Einstein's world is unfortunately a
lost cause; even in Newton's world, if one assumes
finite light-speed, there are difficulties between
the two observers (one moving, one not), or even
two stationary observers.
>
> lol
>
> eleaticus
>
-- #191, ewill3@earthlink.net It's still legal to go .sigless.
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